The value of the integral `int_(0)^(pi)(1)/(a^(2)-2a cos x+1)dx (a gt1)`, is

Options

A. `(pi)/(1-a^(2))`
B. `(pi)/(a^(2)-1)`
C. `(2pi)/(a^(2)-1)`
D. `(2pi)/(1-a^(2))`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{1}{a^2 - 2a \cos x + 1} \, dx \quad (a > 1), \]

we will follow these steps:

### Step 1: Rewrite the integral using the cosine formula
We can express the integral in a more manageable form. We know that:

\[ a^2 - 2a \cos x + 1 = (a - \cos x)^2 + \sin^2 x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\pi} \frac{1}{(a - \cos x)^2 + \sin^2 x} \, dx. \]

### Step 2: Use the substitution \( t = \tan\left(\frac{x}{2}\right) \)
Using the Weierstrass substitution, we have:

\[ \cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2}{1 + t^2} \, dt. \]

When \( x = 0 \), \( t = 0 \) and when \( x = \pi \), \( t \to \infty \). Thus, the limits change accordingly.

### Step 3: Substitute into the integral
Substituting these into the integral, we get:

\[ I = \int_{0}^{\infty} \frac{2}{(a - \frac{1 - t^2}{1 + t^2})^2 + \left(\frac{2t}{1 + t^2}\right)^2} \cdot \frac{2}{1 + t^2} \, dt. \]

### Step 4: Simplify the denominator
Now, we simplify the denominator:

\[ a - \frac{1 - t^2}{1 + t^2} = \frac{a(1 + t^2) - (1 - t^2)}{1 + t^2} = \frac{(a + 1)t^2 + (a - 1)}{1 + t^2}. \]

Thus, the denominator becomes:

\[ \left(\frac{(a + 1)t^2 + (a - 1)}{1 + t^2}\right)^2 + \left(\frac{2t}{1 + t^2}\right)^2. \]

### Step 5: Combine and integrate
After simplifying, we can combine terms and integrate. The integral will yield:

\[ I = \frac{2}{a^2 - 1} \cdot \frac{\pi}{2} = \frac{\pi}{a^2 - 1}. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{a^2 - 1}}. \]